Write the equation of a line that is perpendicular to $y=-0.3x+6$ and that passes through the point $(3,-8)$.
Explanation: Getting started Key idea: The slopes of perpendicular lines are negative reciprocals of each other. Step 1: Find the slope Slope of the given line: ${-0.3}={-\dfrac{3}{10}}$ So, the slope of the perpendicular line: $C{\dfrac{10}{3}}$ Step 2: Substitute the known point into linear equation The perpendicular line will have a slope of $C{\dfrac{10}{3}}$ and pass through the point ${(3,-8)}$. Let's start from the point-slope form of the equation of the perpendicular line, then solve for $y$. [What is the point-slope form?] $\begin{aligned} y-{(-8)} &= C{\dfrac{10}{3}}(x-{3})\\\\\\ y+8 &= C{\dfrac{10}{3}}x -10 \\\\\\ y &=C{\dfrac{10}{3}}x {-18} \end{aligned}$ Answer $y=C{\dfrac{10}{3}}x {-18}$. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $y$ $x$